2 Qs: airplain heading, speed; tan(theta) from vectors

G

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1. An airplane is heading northeast at an airspeed of 700km/hr but a wind is blowing from the west at 60km/hr. In what direction does the plane end up flying and what is its speed relative to the ground?
my work
the wind vector is <60,0>
the airplane vector is <700cos(45'),700sin(45')>
that is as far as I got on that one

2. If VxW=<2,-3,5> and V*W=3 then find tan(theta) where theta is the angle between V and W.
my work
magnitude of VxW is sqrt(38)
sqrt(38)cos(theta)=3
cos(theta)=3/sqrt(38)
sqrt(38)=magnitude of V * magnitude of W sin(theta)
tan=sin/cos
How do I find sin so I can find the tangent? I know the magnitude of the cross product of V and W but not the magnitude of each separately.

Thanks!
 
Flying...

Are you SURE you know which way "West" is? If East and West both are positive, something might be wrong.

What's stopping you from adding the vectors?
 
The wind is coming from the West heading East. The plane is heading Northeast. So, they are both positive. There is nothing keeping me from adding the vectors. But, those numbers just didn't seem right. It would end up being <60+350sqrt2,350sqrt2>
That just didn't seem right to me.
 
Okay, check your answers to see if they make sense. You've got a plane flying at a 45 degree angle and suddenly a wind blows it further east. So the angle it's flying at will be above 45 degrees to the vertical, right?

The vector that describes the plane's velocity is: \(\displaystyle \L\ 700cos(45)i + 700sin(45)j\), which is: \(\displaystyle \L\ 350 \sqrt{2}\ i + 350 \sqrt{2}\ j\).

The vector that describes the wind's velocity is: \(\displaystyle \L\ 60i\).

So, yes, you add them. You're correct. Why don't the numbers seem right? Remember that the speed of the plane flying with the wind is \(\displaystyle \L\ sqrt{(60 + 350\sqrt{2}\)^2 + (350\sqrt{2}\)^2\).

Also, \(\displaystyle \L\ cos(\theta) = \frac{\sqrt{2}\ 350}{x}\\), where x is the speed of the plane with the wind. So that's how you'd find the direction.
 
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